Definition: A graph G = (V, E) is called a tree if G is connected and acyclic. The following theorem captures many important facts about trees.

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1 Tree 1. Trees and their Properties. Spanning trees 3. Minimum Spanning Trees 4. Applications of Minimum Spanning Trees 5. Minimum Spanning Tree Algorithms 1.1 Properties of Trees: Definition: A graph G = (V, E) is called a tree if G is connected and acyclic. The following theorem captures many important facts about trees. Theorem: (Characterizations of trees) Let G = (V, E) be an undirected graph having n vertices and m edges. The following statements are equivalent. 1. G is a tree.. There is a unique path between any two vertices in G. 3. G is connected but G-e is disconnected for every edge e of G. 4. G is connected, and m=n G is acyclic, and m=n G is acyclic but G + xy is cyclic for every x, y V with xy E. Proof: (1) (): Since every tree is connected, there is at least one path between any two vertices in G. Hence, to show that there is a unique path between any two vertices in G, we have to show that there is at most one path between any two vertices in G. We prove this by contradiction. So, assume that there are at least two paths between some pair of vertices, say between x and y. Let P 1 and P be two distinct paths from x to y. By lemma.1, P 1 P contains a cycle. So, G contains a cycle. This contradicts the fact that G is a tree. Hence, there is a unique path between any two vertices in G. () (3): Since, any two vertices in G are connected by a unique path, G is connected. Let xy be any edge in E. Then, P=xy is a path from x to y. So, it must be a unique path from x to y. If we remove xy from G, then there is no path from x to y. Hence, G-xy is disconnected. Since, xy is a arbitrary edge of G, G-e is disconnected for every edge e of G. Hence, G is connected but G-e is disconnected for every edge e of G. (3) (4): By assumption, G is connected. So we need only to show that m=n-1. We prove this by induction. A connected graph with n=1 or n= vertices has n-1 edges. Assume that every graph with fewer than n vertices satisfying (3) also satisfy (4). Suppose that G has n 3 vertices and G satisfies (3), i.e. G is connected but G-e is 1

2 disconnected for every edge e of G. Let e=xy be any edge of G. Now, G-e is disconnected. Now, by lemma 3, G-e has exactly two connected components. Let G 1 and G be the connected components of G. Let n i. and m i, 1 i, be the number of vertices and edges in G i, 1 i. Now, each component satisfies (3), or else G would not satisfy (3). Since, n i < n, i=1,, by induction hypothesis, m i = n i -1, 1 i. So, m = m 1 + m + 1 = (n 1-1) + (n -1) +1 = n-1. So, by induction principle, G has exactly n-1 edges. (4) (5): We have to show that every connected graph G with n vertices and n-1 edges are acyclic. We prove this by induction. For n=1, and 3, it can be easily checked that all connected graph with n vertices and n-1 edges are acyclic. Assume that every connected graph with fewer than n vertices satisfying (4) is acyclic. Let G be a connected graph having n vertices and n-1 edges. Since, G is connected and has n-1 edges, G has a vertex of degree 1. Let x be a vertex of degree 1 in G. Let G =G-x. Now, G is connected and has n-1 vertices and n- edges. So, by induction hypothesis, G is acyclic. Since, x is a degree 1 vertex, x can not be in any cycle of G. Since G =G-x is acyclic, G must be acyclic. So, by induction, every connected graph with n vertices and n-1 edges is acyclic. (5) (6): Suppose that G is acyclic and that m=n-1. Let G i, 1 i k be the connected components of G. Since G is acyclic, G i is acyclic for 1 i k. Hence, each G i, 1 i k is a tree. Let n i and m i, 1 i k, be the number of vertices and edges in G i, 1 i k, k m i i 1 respectively. Since (1) implies (5), m = k = ( n i 1) =n-k. So, k=1. So, G must be a tree. Since (1) implies (), any two vertices in G are connected by a unique path. Thus, adding any edge to G creates a cycle. (6) (1): Suppose that G is acyclic but G +xy is cyclic for every x,y in V with xy E. We must show that G is connected. Let u and v be arbitrary vertices in G. If u and v are not already adjacent, adding the edge uv creates a cycle in which all edges but uv belong to G. Thus, there is a path from u to v and since u and v were chosen arbitrarily, G is connected. Exercises If the maximum degree in a tree T is k, then prove that T has at least k pendant vertices (vertices of degree 1). Is the converse true?. Let T 1 and T be two spanning trees of a connected graph G. If edge e is in T 1 but not in T, prove that there exists another edge f in T but not in T 1 such that (T 1 -e) f and (T -f) e are also spanning trees of G. 3. Prove that in a tree every vertex of degree greater than one is a cut vertex. 4. Prove that a pendant edge in a connected graph G is contained in every spanning tree of G. 5. Prove that an edge e of a connected graph G is a cut edge if and only if e belongs to every spanning tree. 6. Let T be a tree of order m, and let G be a graph with (G) = m-1.then prove that T is isomorphic to some sub graph of G. i 1

3 7. Suppose T is a tree of order n that contains only vertices of degree 1 and 3. Prove that T contains (n-)/ vertices of degree Prove or disprove: if d 1, d, d n is the degree sequence of a tree, then 1, d 1 +1, d, d 3,.d n is the degree sequence of a tree. 9. Let G be a connected weighted graph whose edges have distinct weights. Show that G has a unique minimum spanning tree. 10. Let T be a tree of order n and size m having n i vertices of degree i (i=1, ).show that n 1 = n 3 +n 4 +3n 5 +4n Prove or disprove: if n i denotes the number of vertices of degree i in a tree T, then i n i depend only on the number of vertices in T. 1. Let T be an n vertex tree having one vertex of each degree i, i k; the remaining n-k+1 vertices are leaves. Determine n in terms of k. 13. Draw a weighted connected graph G on 11 vertices having 10 different MSTs. 14. Let e be a minimum cost edge of a weighted connected graph G. Show that e belongs to some MST of G. 15. If e be the only minimum cost edge of G, then e belongs to every MST of G. 16. Describe five applications of MST. 17. Design algorithms for a Tree for each of the following: 1. To find a maximum independent set.. To -color all the vertices of G. 3. To find a path from x to y. 18. Suppose n and d 1, d,,d n,dn+1 are n+1 positive integers such that their sum equals n. Use the pigeon principle to prove that there exists an index i such that d i = 1 and there is an index j such that d j > Use Q18. and Mathematical induction to show that in n is an integer and d 1, n d i i 1 d,,d n are positive integers such that =n-, then there is a tree T n with n vertices whose degrees are d 1, d,, and d n. 0. Characterize all connected graphs with same number of vertices and edges. 3

4 1.: Spanning Tree: Definition : A subgraph T= (V 1, E 1 ) of a graph G = (V,E) is a spanning tree if (i) (ii) T is a tree, and V 1 =V. Theorem 3.1: A graph admits a spanning tree if and only if G is connected. Proof: Necessity: Suppose G admits a spanning tree, say T. We will show that G is connected. Let u, v be any two arbitrary vertices of G. Since, T is a spanning subgraph of G, u and v are vertices of T as well. Since, T is connected, there is a path P(u,v) from u to v in T. As T is a subgraph of G, P(u,v) is also a path in G. Since, u and v are arbitrary vertices of G, there is a path between any two vertices of G. Hence G is connected. Sufficiency: Let G be a connected graph with n vertices and m edges. We construct a spanning tree T of G. Let k=m-n+1. Define G i, 0 i k, recursively, as follows: G if i 0, G i = Gi 1 ei where ei is an edge in some cycle of Gi 1 if 1 i k. Since, G i has exactly n-1+k-i edges, G i is cyclic for each i, 0 i k-1. So, each G i, 0 i k-1, has a cycle. If G i-1 is connected, then G i is also connected, as e i belongs to some cycle of G i-1, 0 i k-1. Hence, Gk is connected and has exactly n-1 edges. So, Gk is a tree. Let T=Gk. Now T is a spanning tree of G. Let (G) denotes the number of distinct spanning trees of a graph. So, by theorem 3.1, (G) 1 for a connected graph. The following graph has exactly three spanning trees. The following theorem gives the number of distinct spanning trees of a complete graph. Theorem 3.: (K n ) = n n- for all n 1. Proof: Since each of K1 and K has exactly one spanning tree (K n ) = n n- for n=1 and. So assume that n 3. Assume that V (K n ) = {1,,n}. Let X be the set of all spanning trees of K n and Y be the set of all sequences a 1,a,,a n- of length n- such that a i {1,,,n}. Note that Y has n n- sequences as each a i, 1 i n- can be selected in n different ways. So, to show that there are n n- spanning trees of K n, it is enough to produce a function f: X Y which is a bijection. Let T be any spanning tree of Kn. Define f(t)= a 1,a,,a n-,a unique sequence of length n- such that a i {1,,,n} for each i, 1 i n-, in the following way. Among all the vertices of degree one, let s 1 be the vertex such s 1 4

5 as an integer is minimum. Let t 1 be the vertex adjacent to s 1 in T. Assign t 1 to s 1. Then, delete the vertex s1 from T. Next, among all the vertices of degree one in T-{s 1 }, let s be the vertex such s as an integer is minimum. Let t be the vertex adjacent to s in T-s 1. Assign t to a. Then, delete the vertex s from T-{s 1 }. Repeat this process until a n- has been defined and a tree with just two vertices remains; the tree T in figure 1, for instance, gives rise to the sequence 6,,,, T 4 3 To show that f is a bijection, we have to prove that (i) no sequence is produced by two different spanning trees of K n and (ii) every sequence of Y is produced by some spanning tree of Kn. We shall achieve both (i) and (ii) by showing that f has an inverse, i.e. we can construct a spanning tree of K n from a sequence a 1,a,, a n- of Y by reversing the process described above of obtaining a sequence of length n- from a spanning tree of K n. Let T be any spanning tree of K n, and let f(t)= a 1,a,,a n-. Then, d(k), the degree of vertex k in T, is equal to the number of times k appears in the sequence a 1,a,,a n-, plus 1. This follows from the observation that when each, but the last, of the edges incident on k is deleted, k appears in the sequence; the last edge may never be deleted, if k is one of the two vertices remaining in the tree, or if it is deleted, k is now the removed leaf, and the adjacent vertex, not k, is included later in the sequence. Thus, if k appears in the sequence then the degree of k in T must be as stated. For example, if f(t) =6,,,,6 is the sequence obtained from a tree T, then d(6)=3, d() = 4, while d(1)=d(3)=d(4)=d(5)=1. So, let f(t)= a 1,a,,a n-. We reconstruct T as follows. Let s 1 be the vertex such that s1 is the least integer in {1,,,n} that does not appear in the sequence a 1,a,,a n-. Join s1 to a1. Then, let s be the vertex such that s is the least integer in {1,,, n}-{s 1 } that does not appear in the sequence a,,a n-. Join s to a. Follow this procedure until s n- is obtained from the sequence a n-. Join s n- to a n-. The tree T is obtained by adding the edge joining the two remaining vertices of N-{s 1, s,,s n- }. For example, the tree T for which f(t)=6,,,,6 is given in figure 1. Thus, we have now established the required one-one correspondence. Hence, the number of distinct spanning tree of K n is n n-. 5

6 1. Minimum Spanning Tree; Let G=(V,E) be a connected weighted graph and C be the cost matrix of G. Let T=(V,E) be a spanning tree of G. The cost of T, denoted C(T), is defined as follows: Definition 3: The cost of a spanning tree T=(V, E ) of a weighted graph G=(V,E) with cost matrix C is defined by C ( e). C(T)= e E Definition 4: A spanning tree T of a weighted connected graph is called a minimum spanning tree if C(T) C(T ) for any other spanning tree T of G. Note that there many be more than one minimum spanning tree of a graph. For example, the graph G in the following figure has exactly two minimum spanning trees G 1 T 1 T 3 A graph G having two minimum spanning trees. 1.4 Minimum Spanning Tree Algorithms In this section, we will discuss two popular algorithms to construct a minimum spanning tree of a weighted connected graph. 6

7 1.4.1 Kruskal s Algorithm As we have seen in the previous section that an acyclic graph with n vertices and n-1 edges is a tree. The first algorithm, known as Kruskal s Algorithm, uses this fact to construct an MST of a connected weighted graph G. We first describe the algorithm informally. First, the algorithm arranges the edges of the graph G in the non-decreasing order of their costs. It starts with the graph T=(V,E ), where E = initially. It then examines each edge for inclusion into the T. If the current edge e under examination does not form a cycle with the so far selected edges, then the edge e is included in T. If e forms a cycle with the so far selected edges, then e is rejected. After the decision of selecting or rejecting the current edge e, the next edge in the list becomes current edge. The algorithm terminates once n-1 edges have been selected or there is no edge left for consideration. The algorithm, thus, maintains acyclicity at each stage of inclusion of edge to T. If the algorithm is successful in adding n-1 edges to T, then T becomes a spanning tree of G. Since the edges are examined for inclusion in the non-decreasing order of their costs, T turns out to be an MST. We, next describe the Kruskal s Algorithm formally. Kruskal s Algorithm Input: A connected Graph G=(V,E) and the cost matrix C of G. Output: A Minimum spanning tree T=(V,E ) of G. Method: Step 1: Sort the edges of G in the non-decreasing order of their costs. Let the sorted list edges be e 1, e,., e m. Step : T=(V, E ), where E =. i=1; count=0; while ( count < n-1 and i < m) { if ( T=(V, E { e i }) is acyclic) { E = E { e i }; count= count+1;} i=i+1; } Next, we prove that the graph T=(V,E ) obtained by Kruskal s algorithm is in fact a minimum spanning tree. Theorem 1.4.1: Kruskal s Algorithm produces a minimum spanning tree in a connected weighted graph. 7

8 Proof: Let G = (V,E) be a connected weighted graph and let T=(V,E ) be the subgraph produced by Kruskal s algorithm for G. Each edge is added to T by Kruskal s algorithm if it does not form a cycle with the already added edges. Hence, T must be acyclic. Next we show that T is connected. If possible T is disconnected and let G 1 = (V 1, E 1 ) be a connected component of T. Let V =V-V 1. Let f 1,f,,f k be the edges of T such that cost(f i ) cost (f i+1 ). Since T is disconnected, k < n-1. Let E =(xy E x V 1 and y V }. Since, G is connected, E is non empty. Let e E be the least cost edge among all edges of E. Let f 1, f,,f i be the edges that have already been selected by Kruskal s Algorithm when the edge e was examined. Since T +e is acyclic, e does not for a cycle with the edges f 1,f,,f i. So, Kruskal s algorithm would have selected the edge e after selecting the edge f i. This is a contradiction to the fact that T is the graph produced by Kruskal s algorithm. Hence, T must be connected. Hence, T is a spanning tree. Next, we show that T has least cost among all spanning trees of G. We prove this by contradiction. Suppose, to the contrary, that T is not a minimum spanning tree. Let E(T)={ f 1,f,,f n-1 } such that w(f i ) w (f i+1 ), 1 i n-. Note that G may have more than one minimum spanning tree. Let T 1 be a minimum spanning tree of G having maximum number of edges in common with T. Let i be the smallest index, 1 i n-1 such that f i is not an edge in T. Such an index exists since T and T 1 are two distinct trees of G. Let T = T 1 +f i. Now, T has a unique cycle C containing f i. Note that C contains at least one edge e 0 such that e 0 is not an edge of T. Let T 3 =T -e 0. Now T 3 is a spanning tree of G and W(T 3 )= W(T 1 )+W(f i )-W(e 0 ). Since, W(T 1 ) W(T 3 ), we have w(e 0 ) w(f i ). By Kruskal s algorithm, fi is an edge of minimum cost such that G[{f 1, f,,f i-1 } {f i }] is acyclic. However, G[{f 1, f,,f i-1 } {e 0 }] is a subgraph of T 1, and hence acyclic. So, w(e 0 )=w(f i ), otherwise Kruskal s algorithm would have chosen e 0 instead of f i. Thus, W(T 3 )=W(T 1 ). Hence, T 3 is a minimum spanning tree. But, T 3 has more edges in common with T than T 1 has with T. This contradicts the fact that T 1 has maximum number of edges in common with T than any other spanning tree of G. Hence, T must a minimum cost spanning tree. We, next, consider another popular algorithm for finding a minimum spanning tree in a connected weighted graph Prim s Algorithm As we have seen above, Kruskal s algorithm starts with the vertex set V and empty edge set and keeps on adding edges maintaining acyclicity throughout. Once n-1 edges are added, it becomes a tree because of the fact that an acyclic graph with n vertices and n-1 edges is a tree. Prim s algorithm adopts a different strategy. It uses the fact that a connected graph with n vertices and n-1 edges is a tree. It starts with vertex set V = {v} where v is any arbitrary vertex of G and E, where E =. It then selects a least cost edge e= xy with x V and y V- V from V to V- V and updates E = E {e} and V = V {y}. It stops when V =V. Thus, it maintains throughout that G =(V, E ) is connected. Once V =V, G becomes a spanning tree of G. We next describe the algorithm formally. 8

9 Prim s Algorithm Input: A connected Graph G=(V,E) and the cost matrix C of G. Output: A Minimum spanning tree T=(V,E ) of G. Method: { Step 1: Let u be any arbitrary vertex of G. T=(V,E ), where V = {u} and E =. Step : while ( V V) { Choose a least cost edge from V to V-V. Let e=xy be a least cost edge such that x V and y V-V. V = V {x}; E = E { e }; } } Next, we show that Prim s algorithm produces a minimum spanning tree of a connected weighted graph. Theorem.3.: Prim s Algorithm produces a minimum spanning tree in a connected weighted graph. Proof: Let G be a connected weighted graph and let T be the subgraph produced by Prim s algorithm. Since, G is connected, T is a spanning tree of G. Next, we show that T is a minimum spanning tree of G. We prove this by method contradiction. Suppose, to the contrary, that T is not a minimum spanning tree of G. Let E(T)={ f 1,f,,f n-1 } such that w(f i ) w (f i+1 ), 1 i n-. Note that G may have more than one minimum spanning tree. Let T 1 be a minimum spanning tree of G having maximum number of edges in common with T. Let i be the smallest index, 1 i n-1 such that f i is not an edge in T. Such an index exists since T and T 1 are two distinct trees of G. For i=1, let U={u}, where u is the first vertex added to V by the Prim s algorithm. If i, then let U be the vertex set of the subgraph induced by the edges f 1,f,,f i-1. Now, f i joins a vertex of U to a vertex of V-U. Let T = T 1 +f i. Now, T has a unique cycle C containing f i. The cycle C contains an edge e 0 that joins a vertex of U to a vertex of V-U. Let T 3 = T 1 +f i e 0. Then, T 3 is a spanning tree of G. Since f i and e 0 are both edges from U to V-U and f i is selected by Prim s algorithm, w(f i ) w(e 0 ). Therefore, w (T 3 ) w (T 1 ). Since, T 1 is a minimum spanning tree, T 3 is also a minimum spanning tree of G. But, T 3 has more edges in common with T than T 1 has with T. This contradicts the choice of T 1. Hence, T must be a minimum spanning Tree. Exercise 1. Let G be a connected weighted graph whose edges have distinct weights. Show that both Prims algorithm as well as Kruskal s algorithm produces the same tree.. Show, for every integer n, there a connected weighted graph having exactly n minimum spanning trees. 9